The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the

Let me try to answer.

You are given that the standard equation of a circle ---> (x−h)^2+(y−k)^2=r^2 where, r=radius of the circle.

Now, you are given another circle with the equation ---> x^2 + y^2 - 8x + 2y = -1 , when you compare this to the standard equation of the circle, you should realize that it would be straightforward to calculate the radius of the circle if you can express x^2 + y^2 - 8x + 2y = -1 in the form of (x−h)^2+(y−k)^2=r^2

In order to make complete squares similar to (x-h)^2 or (y-k)^2, you need to remember that (x+a)^2=x^2+a^2+2ax and (x-a)^2=x^2+a^2-2ax

With these relations in mind, lets go back to x^2 + y^2 - 8x + 2y = -1 with a focus on converting x^2 + y^2 - 8x + 2y to 2 complete squares (1 with x and 1 with y).

x^2 + y^2 - 8x + 2y ---> (x^2-8x+16-16)+(y^2+2y+1-1) = (x-4)^2 -16 + (y+1)^2-1 ---> x^2 + y^2 - 8x + 2y = -1 can thus be written as :

(x-4)^2 -16 + (y+1)^2-1 = -1 --->(x-4)^2 + (y+1)^2 = 16 = 4^2.

NOw you can compare this form to (x−h)^2+(y−k)^2=r^2 to get h = 4, k=-1 and r=radius = 4.

Thus, D is the correct answer.

For questions such as these you need to move backwards from what has been given to you and to modify the given values into a more usable form.

Hope this helps.

Hi Juan,

The trick here is to convert everything in the standard form of equation of a circle, which is:
\((x−h)^2+(y−k)^2=r^2\)

We are given the equation of a circle in some other form (expanded form of the standard equation)

\(x^2+y^2−8x+2y=−1\)
On grouping terms,

\(x^2−8x+y^2+2y=−1\)
From here on, we need to make two perfect squares in x and y

The best way to understand which perfect square to make is by dissecting the coefficient of x and y
8 = 2*1*4, 2 = 2*1*1

We are writing the coefficients in this form, because the coefficient of x in \({(x+a)}^2 = x^2 + 2ax + a^2\) (Eq i) is 2*1*a
\(x^2−2*x*4 + y^2+2*1*y = -1\)

Now we need the constant terms to make these perfect squares.
The constant term in Eq (i) is a

Hence we can write \(x^2−2*1*4x + y^2+2*1*y = -1\) as
\(x^2−2*x*4 + 16 + y^2+2*1*y + 1 = -1 + 16 + 1\) (adding constants on both sides of the equation)
\({(x-4)}^2 + {(y+1)}^2 = 16\)
\({(x-4)}^2 + {(y+1)}^2 = 4^2\)

Hence the radius = 4. Option D

Thank you both, great help.

Bunuel -

I simply factored (x-h)^2 + (y-k)^2 = r^2

[x^2-2xh + h] + [y^2-2yk + y^2] = r^2

If it can be reduced to

x^2 +y^2 -8x +2y -1

Then don't H and K have to be 4 and -1? Unless I'm making some assumption here that wouldn't necessarily apply to some other example- please let me know

If (x-4)^2 + (y - (-1))^2 = r^2
(4-4)^2 + (1 +1)^2 = r^2
(2)^2= r^2
4= r^2
4 = \sqrt{r}
4= r

Yes but don't forget to explain you derived 16 and 1 by taking half of the coefficients 8 and 1 and then squaring them. For example, if we have an algebraic expression that cannot be factored such as x-10x-18=0 - then we complete the square. How?

x-10x __ = 18 (add 18 to both sides or really just put 18 on the other side)
x-10x +25 = 18 + 25 (remember how to get the coefficients? In this equation a= 1 b= 10; take (b/2)^2 and add to both sides
(x-5)(x-5) = 43
x-5 = + or minus 43

VERITAS PREP OFFICIAL SOLUTION:

What should first jump out is that the equation for this particular circle does not match the standard equation of a circle. So you'll need to do some algebra to get to that point.

Your first step should be to factor the x and y terms so that they fit the standard form. Since each term needs to be factored into a squared parenthetical, you should use the common quadratic form \((a−b)^2=a^2−2ab+b^2\) as your guide. Look at the coefficients −8x and +2y, recognizing that they need to fill the −2ab role in that common quadratic equation. This means that you'll factor the x and y terms into:

\((x−4)^2=x^2−8x+16\)
and
\((y−(−1)^2)=y^2+2y+1\)

So now you know that the left-hand side of the standard equation of a circle produces the combination of \(x^2−8x+16+y^2+2y+1\), which equals \(x^2−8x+y^2+2y+17\). So now you know two equations to be true:

\(x^2−8x+y^2+2y+17=r^2\)

and

\(x^2+y^2−8x+2y=−1\)

That second equation can be quickly made to look like the first by adding 1 to both sides:

\(x^2+y^2−8x+2y+1=0\)

Which means that the difference between the two equations is that the left-hand side is 16 apart and the right-hand side is r^2 apart. So to balance, add 16 to the left and r^2 to the right, meaning that r^2=16. This means that the radius is 4, making answer choice D correct.

x^2 + y^2 -8x +2y = -1
x^2 -8x +16 + y^2 + 2y +1 = -1+16+1
(x-4)^2 + (y+1)^2 = 4^2

We can compare the equation with (x-h)^2 + (y-k)^2 = r^2
we get, r=4

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